$H + J^T H J$ has $A + D$ along diagonals and $B - C$ and $C - B$ in antidiagonals, no?

But we want one $n\times n$ thing?

Oh, I'm being dumb.

OK. Usual real form of complex.

Yeah, I think.

So that 2nx2n real matrix is indeed manifesting in GL_n(C)

I'm not getting what you got.

Oh, OK, now I am. You win.

I'm slow.

Whew

No, I am too.

@TedShifrin It's not the whole book it's just the first 10 pages of the book but I gotta say I like the contents

I actually have some physics applications in there, too, @Abdullah, but I doubt you need those. Look at the Harvard book I mentioned, though. It's got a lot of physics and does it all with differential forms :)

Are we trying to legally not-download Ted's books?

@Balarka can testify that he mislearned all of multivariable math from me.

Shhhh.

This is absolutely true I learnt whatever analysis and differential geometry I know because of Ted.

One of my books is legally free in .pdf form on my website (and on the AMS website).

That's not true, a @Balarka.

ok after thinking about this problem and as barlanka pointed the homotopy should be easy

Come on man that's not my name

i drew the path and its just showing a path is homotopic to itsellf just in pieces

balarka****

sorry

THANKS

sorry man

Lol its alrgith

Sigh I misspelled alright

but here is my problem

after some thinking i dont think the homotopy is obvious since i want to find a homotpy between a path and a product of paths which is a branching function. so while the variable $t$ runs i cant have the variable $s$ runs on both the product and the initial path cause that would be something like a tripple product. Now is my homotopy is not a branching function i dont know how to make it give me one function at t=0 and a branching function at t=1

thats how far i got xD

@TedShifrin you mean a course in mathematics for physics?

even if i write down the positive linear maps explicitly its gets weird

Is that worse than my calling you A, a @Balarka? :D

Yes, @Abdullah, for students of physics

i always read it barlanka hahah

Does provide rigorous proofs for its theorems?

It*

@TedShifrin OK, I am very happy now. This seems to say for an oriented hypersurface $\Sigma \subset \Bbb C^n$, the Hermitian form corresponding to the $J$-invariant 2-form $\omega := d\alpha$ where $\alpha = 0$ defines the distribution of maximal complex subspaces of $T\Sigma$ is $\Bbb{II}(X, Y) + \Bbb{II}(JX, JY)$, upto some scalars.

@BalarkaSen dont tell me the answer im just asking are you sure the homotopy is really easy?

A few years ago I could still understand words Balarka said, but now, it's gotten weird

So I guess pseudoconvexity is the same thing as saying the mean curvature along complex tangent lines to $\Sigma$ are positive

Which is totally geometric and I finally get it

@Abdullah: I would have to read carefully, but my recollection is that it's pretty complete. And there are lots of exercises.

@Balarka: And that is cool.

@Krijn: Balarka has moved from algebra to topology to geometry!

Thanks for the all sanity checks

@ManolisLyviakis Yup. Hint: Straightline homotopy

come on man

i thought of that first thing

You still gotta figure out why that's the correct hint

@Balarka: Hörmander's several complex variables book talks about (and proves) equivalence of strictly pseudoconvex and domains of holomorphy.

@Krijn You used to do algebraic number theory, right?

Ah I see

I'll check

Yeah

@TedShifrin OK thanks for the help you made my day

I did some algebraic number theory, some algebraic geometry, and then mixed it up into arithmetic geometry. Also, some class field theory, which is nice

LOL, you're welcome, @Abdullah. Keep me posted :)

Scary stuff @Krijn

It's all very elegant, I think

One man's elegance is the other man's intimidation

And I found out that if you know some elliptic curve stuff you can do lots of cool cryptography

I vaguely knew the ECPP once

Can I make a living out of doing cryptography

That would be cool

National security, @Balarka.

Maybe the NSA will take me or something

Lol

Does India have a variant of it?

Yeah for sure

Yeah, so I actually will be doing a Ph.D. on the subject

And I think industry is willing to take on people in that field

You can easily make a living out of doing cryptography

Nice!

Good night awesome people

What's also cool is that a guy I know started a company in secure multi-party computation

Don't you also need a CS degree for this

So he basically applies advanced cryptography to business settings and that makes a living

Gotcha

A CS degree would have been helpful

I'm not sure actually if more people have a CS degree and learn the maths, or have a maths degree and learn the CS

(specifically for isogeny-based crypto)

Night, @Abdullah.

I know some probability and statistics (part of my course curriculum) so trades/stocks might be a good option for me

I had not considered that

balarka i know i cant add values on X so the straight line homotopy doesnt make sense but i understant the continuous deformation is retract f to a point and then just grow back to the same path in n steps

No, @Manolis, you have to fix the endpoints.

You can't collapse the whole thing to a point

oh right

damn

@BalarkaSen If you want to do trading, you'll get very far if you are arithmetically competent

then im beat

@Krijn Haha, that might be a debatable quality.

@Manolis You have a path f : I -> X and you reparametrize to f o g where g : I -> I, g(0) = 0, g(1) = 1. Just straightline homotope g to identity

Composition of homotopic maps are homotopic (relative to endpoints and all)

The n-step path is just a reparametrization of your original path, really

You go 1/(a_i - a_{i-1}) times faster on each interval [a_{i-1}, a_i]

so u dont calculate a homotopy

H(s,t)

for f and f1*f2...fn

You can of course write down the homotopy given what I have said

ok ill figure out the rest

thanks

np

f is homotopic to is reparametrization

all i need to prove is that the product is a reparametrization

yup

ok munkres writing about positive linear maps kinda got me confused

he just means an endpoint preserving linear reparametrization

yea i get that now xD

i still have trouble visualising the homotopy as a fromula cause the products are branching functions

you dont visualize a formula Manolis

:)

i explained what i mean above

you are right ofc

H(s,t) cant be a branching function

ill have to use the fact that f1*f2 is just f reparametrised so the homotopy will be between f and fog

not f and f1*f2

thanks alot broo goodnight !!!!

Night!

given a map $f:\Bbb R^3 \to \Bbb R^3$ with $f(x,y,z)=(2x,y,3z)$, how do you know which paths all the points take to get to their new destination?

or do they just apparate?

A map f apriori does not come with a homotopy starting at the identity map and ending at --- yeah Ted sniped me

This is sort of a very confusing point, intuitively, because almost all maps I can "see" come with a homotopy (inside some bigger space)

I have to do tricky stuff to see very complicated maps

Oh yeah? How does the identity homotop to the antipodal map on $S^2$?

Haha yeah you win

Oh, I see, you're doing straight-line homotopy in $\Bbb R^3$.

Not interesting.

Well, I see reflection as a rotation one dimension up

Right.

So that's what I have to do here. Draw S^2 as S^1, and do three rotations of the xy-circle in R^3

But I guess I don't actually do that mentally

I never thought about trying to make a function be homotopic to the identity a priori.

I guess flows are the general class of functions where that is natural.

I had to do this to see very complicated singularities for example

Most generic singularities have these sort of factorizations to an immersion one dimension up followed by a projection back

Right.

I think there's some work by Gabai on Stein factorizations of maps between manifolds

where he studies this precisely

are "non" straight line homotopies more interesting or something?

Straightline homotopies are basically useless if you actually want to see what a map does

you can't see a straightline homotopy. its a convenient formula

of course you can see a straightline homotopy

Nah

Draw the straightline homotopy between the antipodal and the identity map S^2 -> S^2 $\subset$ R^3 at time 0, 1/4, 1/2, 3/4, 1

Send the pictures over to me

Good luck

I mean, I guess a simple enough homotopy is just contracting everything to 0 and then expanding back but antipodally this time but this loses all info about the map

Straightline homotopy does that when crossing 0

So that's also another thing

You shouldn't collapse substantial parts of your space in the process of modifying your map to something understandable

I guess what I described is exactly the straightline homotopy. so yeah it's useless for a different reason

at least in this context

But you can imagine drawing the time-slices for a straightline homotopy be terrible for a general map $M \to M$ of manifolds, where you Whitney embed $M$ in $\Bbb R^n$

how do you combine the concept of homotopies with geometry?

That seems like a question from the philosophy land

i can't draw 4 dimensions

like does a specific homotopy a priori carry any geometric information about the space it's traversing?

@Thorgott Drawing $\Bbb C^2$ naively as $\Bbb R^2$ is often surprisingly useful

But here's a question: What's $\{zw = 0\} \cap S^3$ in $\Bbb C^2$? $S^3$ being the unit sphere, and $z, w$ are the two complex coordinates.

This is one of my favorite pictures of all time, I remember the answer being shocking to me when I understood it

It's a möbius string

so in straight line homotopy the points travel along straight lines?

if I may be so egregious